STUDENT SUPPORT MATERIAL

Class X

Mathematics

Session 2016-17

KENDRIYA VIDYALAYA SANGATHAN

NEW DELHI

STUDENT SUPPORT MATERIAL ADVISORS

• Shri Santosh Kumar Mall, IAS, Commissioner, KVS (HQ) New Delhi • Shri. U.N. Khaware, Addl. Commissioner (Academics), KVS (HQ)

CO-ORDINATION TEAM AT KVS (HQ)

• Dr. V. Vijayalakshmi, Joint Commissioner (Acad), KVS (HQ) • Mr. P.V. Sai Ranga Rao, Deputy Commissioner (Acad), KVS (HQ) • Ms. Aprajita, AEO (Acad), KVS (HQ)

CONTENT TEAM • Mr. M.S. Chauhan, D.C., Patna Region (now at RO Chandigarh) • Mr. O. P. Sharma, PGT (Maths), KV Patna Region • Mr. Gaurav Kumar, PGT (Maths), KV Patna Region • Mr. B. K. Jha, TGT (Maths), KV Patna Region • Mr. S. K. Singh, TGT (Maths), KV Patna Region

REVIEW TEAM • Mr. Ajay Gupta, PGT (Maths), KV CRPF, Jharondakalan • Mr. Nitin Maheswari, PGT (Maths), KV Vikaspuri

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INDEX

SL.NO TOPIC SA-1

PART -1 1 Real Numbers

2 Polynomials 3 A pair of linear equations in two variables

4 Triangles

5 Introduction to Trigonometry 6 Statistics

7 Model Question Paper SA-1 PART – 2

8 Activities (Term I)

SLNO TOPIC SA- 2

PART - 1 1 Quadratic Equation

2 Arithmetic Progression

3 Coordinate Geometry 4 Some Applications of Trigonometry

5 Circle 6 Construction

7 Area Related to Circle

8 Surface Area and Volume 9 Probability

10 Model Question paper SA-2 PART – 2

11 Activities (Term II)

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COURSE STRUCTURE

CLASS –X

As per CCE guidelines, the syllabus of Mathematics for class X has been divided term-wise.

The units specified for each term shall be assessed through both formative and summative assessment.

CLASS – X

Term I Term II

FA1 FA2 SA1 FA3 FA4 SA2

(10%) (10%) (30%) (10%) (10%) (30%)

Suggested activities and projects will necessarily be assessed through formative assessment.

SUMMATIVE ASSESSMENT -I

S.NO TOPIC MARKS: 90

SA-I

1 INUMBER SYSTEM Real Numbers

11

2 ALGEBRA Polynomials, pair of linear equations in two variables.

23

3 GEOMETRY Triangles

17

4 TRIGONOMETRY Introduction to trigonometry, trigonometric identity.

22

5 STATISTICS 17

TOTAL 90

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TOPIC WISE ANALYSIS OF EXAMPLES AND QUESTIONS

NCERT TEXT BOOK

Chapters Topics

Number of Questions for revision

Total Questions

from solved

examples

Questions from

exercise

1 Real Number 11 18 29

2 Polynomials 09 08 17

3 Pair of linear equations in two

variables 19 21 40

4 Triangles 14 55 69

5 Introduction to trigonometry 15 27 42

6 Statistics 09 25 34

Total 77 144 231

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DETAILS OF THE CONCEPTS TO BE MASTERED BY EVERY CHILD OF CLASS X WITH

EXERCISE AND EXAMPLES OF NCERT TEXT BOOKS.

SA - I

SYMBOLS USED

S.No

TOPIC CONCEPTS D

EGR

EE OF

IMP

OR

TAN

CE

DIFFICULTY LEVEL

REFERENCES(NCERT BOOK)

TG/LG

DEG

REE

01

Real Number

Euclid’s division Lemma & Algorithm

*** L.G a Example -1,2,3,4 Ex:1.1 Q:1,2,4

Fundamental Theorem of Arithmetic *** L.G a Example -5,7,8 Ex:1.2 Q:4,5

Revisiting Irrational Numbers *** L.G b Example -9,10,11 Ex: 1.3 Q:1.2 Th:1.4

Revisiting Rational Number and their decimal Expansion

** L.G a Ex -1.4 Q:1

02

Polynomials

Meaning of the zero of Polynomial * L.G a Ex -2.1 Q:1

Relationship between zeroes and coefficients of a polynomial

** L.G a Example -2,3 Ex-2.2 Q:1

Forming a quadratic polynomial ** L.G b Ex -2.2 Q:2

Division algorithm for a polynomial * L.G b Ex -2.3 Q:1,2

Finding the zeroes of a polynomial *** L.G a Example: 9 Ex -2.3 Q:1,2,3,4,5 Ex-2.4,3,4,5

03

Pair of Linear

Equations in two

variables

Graphical algebraic representation * L.G b Example:2,3 Ex -3.4 Q:1,3

Consistency of pair of liner equations

** L.G a Ex -3.2 Q:2,4

Graphical method of solution

*** L.G b Example: 4,5 Ex -3.2 Q:7

Algebraic methods of solution a. Substitution method

b. Elimination method

c. Cross multiplication method

d. Equation L.G reducible to pair of

liner equation in two variables

** L.G b

Ex -3.3 Q:1,3 Example-13 Ex:3.4 Q:1,2 Example-15,16 Ex:3.5 Q:1,2,4 Example-19 Ex-3.6

TG/LG is idea identified by termwise error analysis of answers of Q.P. of SA of last three year. * - Important Question a - Low T.G-Teaching Gap ** -Very Important Question b - Average L.G-Learning Gap *** -Very Very Important Question c - Higher

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Q :1(ii),(viii),2 (ii),(iii)

04

Triangles

1) Similarity of Triangles *** T.G C Theo:6.1 Example:1,2,3 Ex:6.2 Q:2,4,6,9,10

2) Criteria for Similarity of Triangles ** T.G C Example:6,7 Ex:6.3 Q:4,5,6,10,13,16

3) Area of Similar Triangles *** L.G B Example:9 The:6.6 Ex:6.4 Q:3,5,6,7

4) Pythagoras Theorem *** L.G b Theo:6.8 & 6.9 Example:10,12,14, Ex:6.5 Q:4,5,6,7,13,14,15,16

05

Introduction to

Trigonometry

1) Trigonometric Ratios * L.G a Ex:8.1 Q:1,2,3,6,8,10

2) Trigonometric ratios of some specific angles

** L.G b Example:10,11 Ex:8.2 Q:1,3

3) Trigonometric ratios of complementary angles

** L.G a Example:14,15 Ex:8.3 Q:2,3,4,6

4) Trigonometric Identities *** L.G b Ex:8.4 Q:5 (iii,v,viii)

06

Statistics

CONCEPT 1 Mean of grouped data

*** L.G a

1. Direct Method * L.G b Example:2 Ex:14.1 Q:1&3

2. Assumed Mean Method * L.G b Ex:14.1 Q:6

3. Step Deviation Method L.G b Ex:14.1 Q:9

CONCEPT 2 *** L.G

Mode of grouped data L.G a Example:5 Ex:14.2 Q:1,5

CONCEPT 3 *** L.G

Median of grouped data L.G a Example:7,8 Ex:14.3 Q1,3,5

CONCEPT 4 ** L.G

Graphical representation of c.f.(give) *** L.G b Example:9 Ex:14.4 Q:1,2,3

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Real Numbers (Key Points)

Real Numbers

Rational Numbers (Q) Irrational Numbers (I)

Natural Numbers (N) (Counting Numbers) (1, 2,3…..)

Whole Numbers (W) (0,1,2,3,4,…)

Integers (Z)

Negative Integers (-1,-2,-3,)

Zero (0)

Positive Integers (1, 2, 3…)

Decimal Form of Real Numbers

Terminating Decimal Non Terminating Non terminating Non Repeating ( 2/5,¾,….) (Rational Numbers)

repeating decimal (Recurring Decimal)

(1/3, 2/7,3/11,…) (Rational Numbers)

(1.010010001…) (Irrational Numbers)

1. Euclid’s Division lemma:-Given Positive integers and b there exist unique integer’s q and r satisfying

a=bq +r, where 0 rd, follow

the steps below:

Step I: Apply Euclid’s division lemma, to c and d, so we find whole numbers, q and r such that c =dq +r,0

Step II: If r=0,d is the HCF of c and d. If r division lemma to d and r. Step III: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF

Note:- Let a and b be positive integers .If a=bq +r, 0≤r

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Theorem: LET x be a rational number whose decimal expansion terminates. Then x can be expressed in the form

Of p/q where are co-prime and the prime factorization of q is of the form of , where n, m are non-negative integers.

Ex.

Theorem: Let 𝑥 =

𝑝

𝑞 be a rational number such that the prime factorization of q is not of the form of

, where n ,m are non-negative integers. Then has a decimal expansion which is none terminating repeating (recurring).

Ex.

Theorem: For any two positive integers a and b, HCF (a,b) XLCM(a,b)=aXb Ex.:4&6; HCF (4,6) =2, LCM(4,6) =12;HCFXLCM=2X12=24

Ans.: aXb=24

LEVEL-I 1. If

𝑝

𝑞 is a rational number (𝑞 ≠ 0).What is the condition on q so that the decimal representation of is

𝑝

𝑞 terminating?

2. Write a rational number between . .

3. The decimal expansion oftherationalno.43/2453 will terminate after how many places of decimal?

4. Find the

5. State whether the number )( + rational or irrational justify.

6. Write one rational and one irrational number lying between 0.25and 0.32.

7. Express 107 in the form of 4q+3 for some positive integer q.

8. Write whether the rational number will have a terminating decimal expansion or a non

Terminating repeating decimal expansion.

9. Show that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+5, where q is some integer.

10. Express 0.2545454…………..As a fraction in simplest form.

LEVEL-II

1. Use Euclid’s division algorithm to find the HCF of 1288 and 575.

2. Check whether 5 x 3 x 11+11 and 5x7+7X3 are composite number and justify.

3. Check whether can end with the digit 0, where n is any natural number.

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4. Given that LCM (26,169) = 338, write HCF (26,169).]

5. Find the HCF and LCM of 6, 72and 120 using the prime factorization method. 6. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or

3m+1 for some integer m.

7. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8

for some integer m.

LEVEL-III .

1. Show that √3 is an irrational number. 2. Show that is an irrational number.

3. Show that square of an odd positive integer is of the form 8m+1, for some integer m.

4. Find the LCM &HCF of 26 and 91 and verify that

5. Prove that ∛7 is irrational.

6. Show that one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer.

7. Find the HCF of 65 & 117 and express it in the form of 65m + 117n.

(PROBLEMS FOR SELF EVALUATION/HOTS)

1. State the fundamental theorem of Arithmetic.

2. Express 2658 as a product of its prime factors.

3. Find the LCM and HCF of 17, 23 and 29.

4. Prove that is not a rational number.

5. Find the largest positive integer that will divide 122, 150 and 115 leaving remainder 5,7 and 11 respectively.

6. Show that there is no positive integer n for which √𝑛 − 1 + √𝑛 + 1 is rational.

7. Using prime factorization method, find the HCF and LCM of 72, 126 and 168. Also show that

HCF X LCM ≠ product of three numbers. 8. Three sets of English, Mathematics and Science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the height of each stack is the same. How many stacks will be there?

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Value Based Questions

Q.1 Aperson wanted to distribute 96 apples and 112 oranges among poor children in an orphanage. He packed all

the fruits in boxes in such a way that each box contains fruits of the same variety, and also every box contains an

equal number of fruits.

(i) Find the maximum number of boxes in which all the fruits can be packed.

(ii) Which concept have you used to find it?

(iii)Which values of this person have been reflected in above situation?

Q.2 A teacher draws the factor tree given in figure and ask the students to find the value of x

without finding the value of y and z.

Shaurya gives the answer x=136

a) Is his answer correct?

b) Give reason for your answer.

c) Which value is depicted in this?

x

2 y

2 z

2 17

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Answer Level-I

1. q is of the form 2n .5m , where m and n are non-

negative integers.

2. 1.5

3. After 4 places of decimal.

4. 19000

5. Rational number

6. One rational number=26/100, one irrational

no.=0.27010010001………

7. 4 X 26+3

8. Terminating

10.14/55

Level-II

1.23

2. Composite number

3. No, 6n cannot end with the digit 0.

4.13

5. HCF=6 , LCM = 360

Level-III 4. LCM= 182 ,HCF = 13

7. m = 2 and n = -1.

Problems for self-evaluation 1. See textbook.

2. 2658 = 2 X 3 x 443

3. HCF = 1 , LCM = 11339

5. 13 8. Total no. of stacks = 14

Value based Questions

1. (i)No. of boxes = 16

(ii)Number System & HCF (iii)The person is kind hearted and of helping attitude. 2. (a) Yes, his answer is correct.

(b) Z =2 X 17 = 34, Y = 2 X 34 = 68, X = 2 x 68 = 136 (c) Knowledge of prime factorization.

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Polynomial Polynomial An expression of the form p(x) = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥

2 + ⋯ … … … … . +𝑎𝑛𝑥𝑛 where 𝑎𝑛 ≠ 0 is

called a polynomial in one variable x of degree n, where; 𝑎0, 𝑎1,𝑎2 … … … … … . . 𝑎𝑛 are constants and they are called the coefficients of 𝑥0, 𝑥, 𝑥

2 … … … 𝑥𝑛 . Each power of x is a non-negative integer. Eg: -2𝑥2 − 5𝑥 + 1 is a polynomial of degree 2

Note: √𝑥 + 3 is not a polynomial

A polynomial p(𝑥) = 𝑎𝑥 + 𝑏 of degree 1 is called a linear polynomial Eg: 5𝑥 − 3, 2𝑥 etc

A polynomial p(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 of degree 2 is called a quadratic polynomial Eg:

2𝑥2 + 𝑥 − 1

A polynomial 𝑝(𝑥) = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑 of degree 3 is called a cubic polynomial.

Eg: √3𝑥3 − 𝑥 + √5, 𝑥3 − 1 etc Zeroes of a polynomial: A real number k is called a zero of polynomial p(x) if p(k)=0. If the graph of y= p(x) intersects the X-axis at n times, the number of zeroes of y= p(x) is n.

A linear polynomial has only one zero.

A quadratic polynomial has two zeroes.

A cubic polynomial has three zeroes.

Graphs of different types of polynomials:

Linear polynomial:- The graph of a linear polynomial ax+b is a straight line, intersecting

X- axis at one point

Quadratic polynomial:-

(i) Graph of a quadratic polynomial 𝑝(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 is a parabola open upwards like U, if a>0 & intersects x-axis at maximum two distinct points.

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(ii) Graph of a quadratic polynomial p(x)=𝑎𝑥2 + 𝑏𝑥 + 𝑐 is a parabola open downwards like ∩ if a

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𝛼 + 𝛽 + 𝛾 = −𝑏

𝑎

𝛼𝛽 + 𝛽𝛾 + 𝛾𝛼 =𝑐

𝑎

𝛼𝛽𝛾 =−𝑑

𝑎

If 𝛼, 𝛽&𝛾 are zeroes of a cubic polynomial p(x),

𝑝(𝑥) = 𝑥3 − (𝛼 + 𝛽 + 𝛾)𝑥2 + (𝛼𝛽 + 𝛽𝛾 + 𝛾𝛼)𝑥 − 𝛼𝛽𝛾

Division algorithm for polynomials: If p(x) and g(x) are any two polynomials with g(x)≠ 0, then we have polynomials q(x) and r(x) such that P(x) = g(x)× 𝑞(𝑥) + 𝑟(𝑥), where r(x) =0 or degree of r(x) 0

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Case- 3 When Polynomial 𝑎𝑥2 + 𝑏𝑥 + 𝑐 is not factorizable. In this case, the curve doesn’t cut or touches X-axis

Level –I

1. Find the value of zeroes of the polynomials p(x) as shown in the graph and hence find

the polynomial.(CBSE 2014-15).

2. Let α and β are the zeroes of a quadratic polynomial 2𝑥2 − 5𝑥 − 6 then form a

quadratic polynomial whose zeroes are 𝛼 + 𝛽 𝑎𝑛𝑑 𝛼𝛽. (CBSE 2011)

(i) a>0 (ii) a

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3. Check whether 𝑥2 + 3𝑥 + 1 is a factor of 3𝑥4 + 5𝑥3 − 7𝑥2 + 2𝑥 + 2?

(CBSE 2010) 4. Can (x-7) be the remainder on division of a polynomial 𝑝(𝑥) 𝑏𝑦 (7𝑥 + 2)? Justify your

answer(CBSE 2010)

5. What must be subtracted from the polynomial 𝑓(𝑥) = 𝑥4 + 2𝑥3 − 13𝑥2 − 12𝑥 + 21,

so that the resulting polynomial is exactly divisible by 𝑥2 − 4𝑥 + 3? (CBSE 2013)

6. Write the degree of zero polynomial?

7. Find the zeroes of a quadratic polynomial 6𝑥2 − 7𝑥 − 3 and verify the relationship

between the zeroes and the coefficients? (CBSE 2014-15

8. Find the quadratic polynomial sum of whose zeroes is 2√3 and their product is 2?(CBSE

2008)

Level II

9. If the sum of squares of the zeroes of the polynomials 6𝑥2 + 𝑥 + 𝑘 is 25

36. find the value

of k?( CBSE 2014-15)

10. If one zero of the quadratic polynomial f(x)= 4𝑥2 − 8𝑘𝑥 − 9 is negative of the other,

then find the value of k?(CBSE 2014-15)

11. Find the values of k for which the quadratic equation 9𝑥2 − 3𝑘𝑥 + 𝑘 = 0 has equal

roots. (CBSE 2014)

12. On dividing 3𝑥3 − 2𝑥2 + 5𝑥 + 5 by the polynomial p(x), the quotient and remainder

are 𝑥2 − 𝑥 + 2 and−7 respectively. Find p(x)?(CBSE 2013)

13. Find all the zeroes of the polynomial 𝑥4 + 𝑥3 − 9𝑥2 − 3𝑥 + 18, if two of its zeroes are

√3 𝑎𝑛𝑑√−3. (CBSE 2010,13)

14. If 𝛼 , 𝛽 are zeroes of the quadratic polynomial 𝑝(𝑥) = 𝑥2 − (𝑘 − 6)𝑥 + (2𝑘 + 1). Find

the value of k if 𝛼 + 𝛽 = 𝛼𝛽. (CBSE 2010)

15. If the zeroes of the polynomial 𝑥2 − 5𝑥 + 𝑘 are the reciprocal of each other, then find

the value of K? (CBSE 2011)

16. If α and β are zeroes of the quadratic polynomial 𝑥2 − 6𝑥 + 𝑎, find the value of′𝑎′. If

3𝛼 + 2𝛽 = 20.(CBSE 2010)

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LEVEL III

17. On dividing 3𝑥3 + 4𝑥2 + 5𝑥 − 13 by a polynomial g(x), the quotient and remainder are

3𝑥 + 10 and 16𝑥 − 43 respectively. Find the polynomial g(x). (CBSE 14-15)

18. If -5 is a root of quadratic equation 2𝑥2 + 𝑝𝑥 − 15 = 0 and the quadratic equation

𝑝(𝑥2 + 𝑥)𝑘 = 0 has equal roots, find the value of k. (CBSE 2106)

19. If 𝛼, 𝛽 𝑎𝑛𝑑 𝛾 are zeroes of the polynomial 6𝑥3 + 3𝑥2 − 5𝑥 + 1, then find the values of

𝛼−1 + 𝛽−1 + 𝛾−1. (CBSE 2010)

20. Form a cubic polynomial whose zeroes are 3, 2 and -1. Hence find

(i) Sum of its zeroes

(ii) Sum of the product, taken two at a time

(iii) Product of its zero.

(SELF EVALUATION QUESTIONS) 21. Find the number of zeroes of p(x) in each case, for some polynomials p(x).

22. If 𝛼 𝑎𝑛𝑑𝛽 are the zeroes of the equation 6𝑥2 + 𝑥 − 2 = 0, find 𝛼

𝛽+

𝛽

𝛼

23. If one of the zeroes of the polynomial 2𝑥2 + 𝑝𝑥 + 4 = 0 is 2, find the other zero, also

find the value of p

24. If one zero of the polynomial (𝑎2 + 9)𝑥2 + 13𝑥 + 6𝑎 is reciprocal of the other. Find

the value of a. (All India)

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Value Based Questions

25. If 𝛼 be the number of person who take junk food, 𝛽 be the person who take food at

home and α and β be the zeroes of quadratic polynomial 𝑓(𝑥) = 𝑥2 − 3𝑥 + 2, then

find a quadratic polynomial whose zeroes are 1

2𝛼+𝛽 𝑎𝑛𝑑

1

2𝛽+𝛼 , which way of taking

food you prefer and why?

26. If the number of apples and mangoes are the zeroes of the polynomial 3𝑥2 = 8𝑥 −

2𝑘 + 1 and the number of apples is 7 times the number of mangoes, then find the

number of zeroes and value of k. What are benefits of fruits in our daily life?

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Pair of Linear Equations in Two Variables

(Key Points)

An equation of the form ax + by + c = 0, where a, b, c are real nos. (a 0, b 0) i.e (a2+b2 ≠ 0) is called

a linear equation in two variables x and y.

Ex : (i) x – 5y + 2 =0

(ii) 3

2 x – y =1

The general form for a pair of linear equations in two variables x and y is

a1x + b1y + c1 = 0 a2x + b2y + c2 = 0

Where a1, b1, c1, a2, b2, c2 are all real nos and a1 0, b1 0, a2 0, b2 0.

Examples

Graphical representation of a pair of linear equations in two variables:

a1x + b1y + c1 = 0 a2x + b2y + c2 = 0

(i) Will represent intersecting lines if

I.e. unique solution. And these types of equations are called consistent pair of linear equations. Ex: x – 2y = 0 3x + 4y – 20 = 0

(ii) will represent overlapping or coincident lines if

i.e. Infinitely many solutions, consistent or dependent pair of linear equations Ex: 2x + 3y – 9 = 0 4x + 6y – 18 = 0

Co-ordinates of the point of intersection gives the solution of the equations.

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(iii) will represent parallel lines if

i.e. no solution and called inconsistent pair of linear equations. Ex: x + 2y – 4 = 0 2x + 4y – 12 = 0

• Algebraic methods of solving a pair of linear equations:

(i) Substitution method

(ii) Elimination Method

(iii) Cross multiplication method

Level - I

1. Find the value of ‘a’ so that the point(2,9) lies on the line represented by ax-3y=5

2. Find the value of k so that the lines 2x – 3y = 9 and kx-9y =18 will be parallel.

3. Find the value of k for which x + 2y =5, 3x+ky+15=0 is inconsistent

4. Check whether given pair of lines is consistent or not 5x – 1 = 2y, y = +

5. Determine the value of ‘a’ if the system of linear equations 3x+2y -4 =0 and ax – y – 3 = 0 will

represent intersecting lines.

6. Write any one equation of the line which is parallel to 2x – 3y =5

7. Find the point of intersection of line -3x + 7y =3 with x-axis

8. For what value of k the following pair has infinite number of solutions.

(k-3)x + 3y = k

k(x+y)=12

9. Write the condition sothat a1x + b1y = c1 and a2x + b2y = c2 have unique solution.

The graph is Coincident lines,

Parallel lines, no solution.

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Level - II

1. 5 pencils and 7pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. Find the

cost of one pencil and that of one pen.

2. Solve the equations:

3x – y = 3

7x + 2y = 20

3. Find the fraction which becomes to 2/3 when the numerator is increased by 2 and equal to 4/7 when

the denominator is increased by 4

4. Solve the equation:

px + qy = p – q

qx – py = p + q

5. Solve the equation using the method of substitution:

6. Solve the equations:

Where, x

7. Solve the equations by using the method of cross multiplication:

5x + 12y =7

Level - III

1. Draw the graph of the equations

4x – y = 4

4x + y = 12

Determine the vertices of the triangle formed by the lines representing these equations and the x-

axis. Shade the triangular region so formed

2. Solve Graphically

x – y = -1 and

3x + 2y = 12

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Calculate the area bounded by these lines and the x- axis,

3. Solve :- for u & v

4u – v = 14uv

3u + 2v = 16uv where u≠0, v≠ 0

4. Ritu can row downstream 20 km in 2 hr , and upstream 4 km in 2 hr . Find her speed of rowing in still

water and the speed of the current. (HOTS)

5. In a , = 3∠B = 2 (∠A +∠B ) find the these angle. (HOTS)

6. 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14

days. Find the time taken by 1 man alone and that by one boy alone to finish the work. (HOTS)

7. Find the value of K for which the system of linear equations 2x+5y = 3, (k +1 )x + 2(k + 2) y = 2K will

have infinite number of solutions. (HOTS)

SELF EVALUTION

1. Solve for x and y:

x + y = a + b

ax – by=

2. For what value of k will the equation x +5y-7=0 and 4x +20y +k=0 represent coincident lines?

3. Solve graphically: 3x +y +1=0

2x -3y +8=0

4. The sum of digits of a two digit number is 9. If 27is subtracted from the number, the digits are

reversed. Find the number.

5. Draw the graph of x + 2y – 7 =0 and 2x – y -4 = 0. Shade the area bounded by these lines and Y-axis.

6. Students of a class are made to stand in rows. If one student is extra in a row, there would be 2 rows

less. If one student is less in a row there would be 3 rows more. Find the number of the students in the

class.

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7. A man travels 370 km partly by train and remaining by car. If he covers 250 km by train and the rest by

the car it takes him 4 hours, but if he travels 130 km by train and the rest by car, he takes 18 minutes

longer. Find the speed of the train and that of the car.

8. Given linear equation 2x +3y-8=0, write another linear equation such that the geometrical

representation of the pair so formed is (i) intersecting lines, (ii) Parallel Lines.

9. Solve for x and y.

(a-b)x +(a+b)y = a2 - 2ab – b2

(a+b)(x+y) = a2+b2(CBSE 2004, ’07C, ’08)

10. The sum of two numbers is 8 and the sum of their reciprocal is 8/15. Find the numbers.

(CBSE 2009)

Value Based Questions

Q1. The owner of a taxi cab company decides to run all the cars he has on CNG fuel instead of

petrol/diesel. The car hire charges in city comprises of fixed charges together with the charge for the

distance covered. For a journey of 12km, the charge paid Rs.89 and for a journey of 20 km, the charge

paid is Rs. 145.

i. What will a person have to pay for travelling a distance of 30 km?

ii. Which concept has been used to find it?

iii. Which values of the owner have been depicted here?

Q2.Riya decides to use public transport to cover a distance of 300 km. She travels this distance partly

by train and remaining by bus. She takes 4 hours if she travels 60km by bus and the remaining by

train. If she travels 100 km by bus and the remaining by train, she takes 10 minutes more.

i. Find speed of train and bus separately.

ii. Which concept has been used to solve the above problem?

iii. Which values of Riya have been depicted here?

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ANSWER

LEVEL-I

Q1.a= 16 Q2.k= 6 Q3.k= 6 Q4.Consistent

Q5. a

Q6. (May be another solution also) Q7.(-1, 0) Q8.k= 6

Q9.

LEVEL-II

Q1.: Cost of one pencil = Rs. 3

Cost of one pen = Rs. 5

Q2.x=2, y=3 Q3.28/45 Q4.x = 1, y = -1

Q5.

Q6.

Q7.

LEVEL-III

Q1.(2,4)(1,0)(3,0) Q2.x = 2, y = 3 and area = 7.5 unit 2

Q3.u = ½ , v = ¼ Q4. Speed of the rowing in still water = 6 km/hr

Speed of the current = 4 km/hr .

Q5. ∠A = 200,∠B = 400, ∠C = 1200.

Q6.: One man can finish work in 140 days.

One boy can finish work in 280 days.

Q7.K = 3

SELF EVALUATION Q1.X=a,y=b Q2.K=-28 Q3.X= -1, y=2 Q4.63 Q6.60

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Q7.Speed of the train=100km/h, speed of the car=80km/h Q8.(i) 4x-3y-8 =0 (may be another equation also) (ii) 4x+6y+16 =0 (may be another equation also) Q9.X= a+b,y = -2ab/(a+b) Q10.3,5

VALUE BASED QUESTIONS

Q1.(i)Rs.215,(ii)A pair of linear equations in two variables has been used to find it.

(iii) Awareness of environment.

Q2. (i) The speed of the train = 80 km/h, the speed of the bus = 60km/h

(ii) A pair of linear equations in two variables has been used.

(iii)Controlling the pollution of the environment.

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Triangles

Key Points

Similar Figures: Two figures having similar shapes (size may or may not same), called Similar figures. Examples: (a) & (b) & (c) & A pair of Circles A pair of squares A pair of Equ. Triangles

Pairs of all regular polygons, containing equal number of sides are examples of Similar

Figures.

Similar Triangles: Two Triangles are said to be similar if

(a) Their corresponding angles are equal ( also called Equiangular Triangles)

(b) Ratio of their corresponding sides are equal/proportional

All congruent figures are similar but similar figures may /may not congruent

Conditions for similarity of two Triangles

(a) AAA criterion/A-A corollary

(b) SAS similarity criterion

(c) SSS similarity criterion (where ‘S’ stands for ratio of corresponding sides of two

Triangles)

Important Theorems of the topicTriangles

(a) Basic Proportionality Theorem (B.P.T.)/Thale’s Theorem

(b) Converse of B.P.T.

(c) Area related theorem of Similar Triangles

(d) Pythagoras Theorem

(e) Converse of Pythagoras Theorem

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Level I

(1) In the figure XY ∕∕ QR , PQ/ XQ = 7/3 and PR =6.3cm then find YR

(2) If ∆ABC ~ ∆ DEF and their areas be 64cm2& 121cm2 respectively , then find BC if EF =15.4 cm

(3) ABC is an isosceles ∆ ,right angled at C then prove that AB2 = 2AC2

(4) If ∆ABC ~ ∆ DEF, ∟A=460, ∟E= 620 then the measure of ∟C=720. Is it true? Give reason.

(5) The ratio of the corresponding sides of two similar triangles is 16:25 then find the ratio of their

perimeters.

(6) A man goes 24 km in due east and then He goes 10 km in due north. How far is He from the starting

Point?

(7) The length of the diagonals of a rhombus is 16cm & 12cm respectively then find the perimeter of

the rhombus.

(8) In the figure LM ∕∕CB and LN ∕∕ CD then prove that AM/AB = AN /AD

(9) Which one is the sides of a right angled triangles among the following (a) 6cm,8cm & 11cm (b)

3cm,4cm & 6cm (c) 5cm , 12cm & 13cm

Level II

(1) In the figure ABD is a triangle right angled at A and AC is perpendicular to BD then show that AC2=

BC x DC

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(2) Two poles of height 10m & 15 m stand vertically on a plane ground. If the distance between their

feet is 5√3m then find the distance between their tops.

(3) D & E are the points on the sides AB & AC of ∆ABC, as shown in the figure. If ∟B = ∟AED then

show that ∆ABC ~∆AED

(4) In the adjoining figure AB ∕∕ DC and diagonal AC & BD intersect at point O. If AO = (3x-1)cm , OB=

(2x+1)cm, OC=(5x-3 )cm and OD=( 6x-5)cm then find the value of x.

(5) In the figure D &E trisect BC. Prove that 8AE2= 3AC2+ 5AD2

(6) In the figure OA/OC = OD /OB then prove that ∟A= ∟C

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(7) Using converse of B.P.T. prove that the line joining the mid points of any two sides of a triangle is

parallel to the third side of the triangle.

(8) In the given figure ∆ABC &∆ DBC are on the same base BC . if AD intersect BC at O then prove that

ar(∆ABC)/ar(∆ DBC) = AO/DO

Level III

(1) A point O is in the interior of a rectangle ABCD, is joined with each of the vertices A, B, C & D. Prove

that OA2 +OC2 = OB2+OD2

(2) In an equilateral triangle ABC, D is a point on the base BC such that BD= 1/3 BC ,then show that

9AD2= 7AB2

(3) Prove that in a rhombus, sum of squares of the sides is equal to the sum of the squares of its

diagonals

(4) In the adjoining figure ABCD is a parallelogram. Through the midpoint M of the side CD, a line is

drawn which cuts diagonal AC at L and AD produced at E. Prove that EL =2BL

(5) ABC & DBC are two triangles on the same base BC and on the same side of BC with ∟A = ∟D =900. If

CA & BD meet each other at E then show that AE x EC = BE x ED

(6) ABC is a Triangle, right angle at C and p is the length of the perpendicular drawn from C to AB. By

expressing the area of the triangle in two ways show that (i) pc =ab (ii) 1 /p2 = 1/a2 +1/b2

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(7) Prove that the ratio of the areas of two similar triangles is equal to the ratio of their corresponding

sides.

(8) In the figure AB|| DE and BD|| EF. Prove that DC2= CF x AC

Self-Evaluation Questions including Board Questions &Value Based Questions

(1) Find the value of x for which DE ||BC in the adjoining figure

(2) In an equilateral triangle prove that three times the square of one side is equal to four times the

square of one of its altitude.

(3) The perpendicular from A on the side BC of a triangle ABC intersect BC at D such that DB = 3CD.

Prove that 2AB2= 2AC2+ BC2

(4) In the adjoining figure P is the midpoint of BC and Q is the midpoint of AP. If BQ when produced

meets AC at R ,then prove that RA = 1/3 CA

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(5) BL and CM are medians of triangle ABC , right angled at A then prove that 4(BL2+CM2)= 5BC2

(6) In ∆ABC if AB =6√3cm , AC =12cm and BC=6cm then show that ∟B =900

(7) In the adjoining figure ∠QRP =900, ∠PMR=900,QR =26cm, PM= 8cm and MR =6cm then find the

area of ∆PQR

(8) If the ratio of the corresponding sides of two similar triangles is 2:3 then find the ratio of their

corresponding altitudes.

(9) In the adjoining figure ABC is a ∆ right angled at C. P& Q are the points on the sides CA & CB

respectively which divides these sides in the ratio 2:1, then prove that 9(AQ2 + BP2 ) = 13 AB2

(10) In the adjoining figure AB || PQ ||CD, AB =x unit, CD= y unit & PQ = z unit then prove that 1/x +1/ y = 1/z

(11)State and prove Pythagoras theorem. Using this theorem find the distance between the tops of two vertical poles of height 12m & 18m respectively fixed at a distance of 8m apart from each other.

A

C B

P

Q

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(12) in the adjoining figure DEFG is a square & ∟BAC= 900 then prove that (a) ∆AGF ~ ∆ DBG (B) ∆ AGF ~∆ EFC (C) ∆ DBG ~∆ EFC (D)DE2 = BD X EC

(13) A man steadily goes 4 m due east and then 3m due north .Find

(a) Distance from initial point to last point. (b) What mathematical concept is used in this problem? (c) What is its value?

A

C D

F

B

G

E

P a g e 34 | 118

Solutions

Level I

(1) By B.P.T. PQ/XQ=PR/YR⇨ 7/3=6.3/YR ⇨ YR= 3x6.3/7=2.7

So YR=2.7cm

(2) By theorem Ar of ∆ABC/Ar of ∆DEF= BC2/15.42

⇨64/121 = BC2/15.42⇨solving BC = 11.2 cm

(3) By Pythagoras theorem AB2= AC2+BC2 ⇨AB2= AC2+AC2(given that AC=BC) So AB2=2AC2

(4) ∆ABC~∆DEF ⇨ ∠A=∠D=460 , ∠B =∠ E=620 so ∠C =180-(46+62)=720

So it is true. (5) Let ∆ABC~∆DEF

then AB/DE= BC/EF=AC/DF= perimeter of ∆ABC/Perimeter of ∆DEF ⇨AB/DE=perimeter of ∆ABC/Perimeter of ∆DEF So perimeter of ∆ABC/Perimeter of ∆DEF=16:25

(6) By Pythagoras theorem , Distance =√ 242+102

On Solving , distance =26km (7) In ∆AOD, by Pythagoras theorem AD=√62+82

⇨AD= 10cm So perimeter of Rhombus = 4x10cm = 40cm

(8) In ∆ABC ,LM//BC so by BPT AM/AB=AL/AC------(i)

Similarly in ∆ACD , LN//DC , so by BPT AN/AD = AL/AC-----------(ii) Comparing results I &ii we get AM/AB= AN/AD

Using Pythagoras thermo ,finding the value of p2+b2&h2 separately in each case , it comes equal in case of c where p2+b2 comes equal to h2

So sides given in question c is the sides of right triangle

Level II

(1)In ∆ABDABC ∠2+∠3=900

⇨∠1+∠2=∠2+∠3

⇨ ∠1=∠3 ∆ACD~∆BCA ⇨AC/BC= CD/AC So AC2=BC x CD

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(2)Using Pythagoras theorem

Distance between their tops = √52+(5√3)2

√ 25 + 75 Distance between their tops= 10m (3)In ∆AED&∆ABC ∠AED=∠ABC(given) ∠A=∠A(common) By AA corollary∆ ABC ~ ∆AED (4)Diagonals of a trapezium divide each proportionally So AO/OC =BO/OD 3x-1/5x-3= 2x+1/6x-5 ⇨8x2-20x+8=0 Solving we get x=2 &1/2(na) So x=2 (5) BD=DE=EC =P(let) BE=2P &BC=3P In Rt ∆ABD ,AD2= AB2+BD2 =AB2+p2 In Rt ∆ABE, AE2= AB2+BE2 =AB2+(2p)2 =AB2+4p2 In Rt ∆ABC,AC2=AB2+BC2 =AB2+(3p)2 =AB2+9p2 Now taking RHS 3AC2+5AD2 =3(AB2+9p2)+5(AB2+p2) =8AB2+32p2 =8(AB2+4p2) =8AE2 =LHS (6)OA/OC=OD/OB(given) ⇨OA/OD=OC/OB &∠AOD=∠BOC(v.o.∠s) By SAS similarity condition ∆AOD~∆COB ⇨∠A=∠C (7)

Given that AD/DE=1 &AE/EC=1(as D &E are mid points of the sides AB & AC) ⇨AD/DB =AE/EC By converse of BPT DE//BC

A

B C

D E

P a g e 36 | 118

(8) We draw perpendiculars AM & DN as shown .∆DON~∆AOM(by AA corollary) DN/AM =OD/OA⇨AM/DN=OA/OD-------(i) Ar of ∆ABC/Ar of ∆DBC =(1/2xBCx AM)/(1/2x BC x DN) =AM/DN Ar of ∆ABC/Ar of ∆DBC =AO/OD(from (i))

Level III

(1) We draw PQ ||BC through Pt. O⇨BPQC & APQD are rectangles.

In Rt ∆OPB , by Pythagoras theorem OB2=BP2+OP2-------(i) In Rt ∆OQD ,OD2=OQ2+DQ2----------(ii) In Rt ∆OQC ,OC2=OQ2+CQ2--------(iii) In Rt ∆OAP, OA2=AP2+OP2-------------(iv) On adding (i) &(ii) OB2+OD2=BP2+OP2+OQ2+PQ2 =CQ2+OP2+OQ2+AP2(BP=CQ & DA =AP) =CQ2+OQ2+OP2+AP2 So OB2+OD2 =OC2+OD2 (2)

We draw AE perpendicular to BC & AD is joined. Then BD = BC/3 , DC =2BC/3 & BE=EC =BC/2 In Rt. ∆ADE,AD2=AE2+DE2 =AE2+(BE-BD)2 =AE2+BE2+BD2-2.BE.BD = AB2+ (BC/3)2-2.BC/2.BC/3 =AB2+BC2/9-BC2/3 =(9AB2+BC2-3BC2)/9 -=(9AB2+AB2-3AB2)/9( Given AB=BC=AC) =7AB2/9 ⇨9AD2=7AB2 (3)In Rt.∆AOB,AB2=OA2+OB2 =(AC/2)2+(BD/2)2 4AB2= AC2+BD2---------------(I) Similarly 4BC2=AC2+BD2----------------(II) 4CD2= AC2+BD2--------------(III) 4AD2=AC2+BD2-------------(IV) Adding these results 4(AB2+BC2+CD2+AD2) =4(AC2+BD2) ⇨(AB2+BC2+CD2+AD2) =(AC2+BD2)

M

N

A

B C

D E

P a g e 37 | 118

(4)∆BMC ≅∆EDM(by ASA criterion) ⇨by cpct DE=BC & AD =BC (opp. sides of //gm) Adding above results AD+DE=BC+BC ⇨AE =2BC Now ∆AEL~∆CBL (By AA corollary) EL/BL=AE/BC⇨EL/BL =2BC/BC ⇨EL =2BL (5)∆AEB~∆DEC(AA corollary) AE/DE=EB/EC ⇨AE X EC= BE X ED (6)Ar of ∆ABC=1/2x AB X DC =1/2 X c Xp = pc/2 Again Ar of ∆ABC= ½ x AC X BC =1/2 x b x a = ab/2 Comparing above two areas ab/2= pc/2 ⇨ab=pc Now in Rt ∆ABC,AB2=BC2+AC2 c2=a2+b2 (ab/p)2= a2+b2)(ab=pc⇨c=ab/p) a2b2/p2=a2+b2 1/p2=a2+b2/a2b2 1/p2= 1/a2 +1/b2

(7) Theorem question, as proved (8) In ∆ABC,AB //DE , by BPT AC/DC BC/CE--------(i) In ∆DBC, EF//BD, by BPT DC/CF = BC/EC -----------(ii) Comparing (i) &(ii) AC/DC=DC/CF ⇨DC2=AC X CF

Self-Evaluation Questions

(1) A/Q AD/DB = AE/EC (by BPT)

⇨x/3x+1= x+3/3x+11 ⇨3x2+11=3x2+9x+x+3 So x=3

(2)

In ∆ABD,AB2=AD2+BD2 = AD2+(BC/2)2(AB=BC=AC)

A D

B C

E

A

B C

D

P a g e 38 | 118

-= AD2+AB2/4 4AB2=4AD2+AB2 4AB2-AB2= 4AD2 3AB2=4AD2 (3), BC =4CD ⇨CD = BC/4 ⇨ BD =3CD= 3BC/4 ---------(i) In ∆ABD ,AB2=AD2+BD2-------(ii) In ∆ACD, AC2= AD2+CD2------(iii) Now AB2-AC2= BD2=CD2 = 9BC2/16- BC2/16= BC2/2 2( AB2-AC2) =BC2 2AB2=2AC2+BC2 (4) we draw PS||BR In triangle RBC, P is the mid point of BC and PS||BR RS=CS [Mid point theorem] ……………………..(1) In ∆ APS, PS||BR ie PS||QR and Q is the mid point of AP So AR=RS………………………..[II](Mid point theorem) From results (I)&(II) AR=RS=CS So AR =1/3AC (5) In ∆ABL BL2=AB2+AL2 4BL2=4AB2+4AL2 =4AB2+(2AL)2 4BL2=4AB2+AC2----(i) In ∆ACM 4CM2=4AC2+AB2----(ii) On adding 4BL2+4CM2=4AB2+AC2+4AC2+AB2 =5AB2+5AC2 =5(AC2+AB2) =5BC2 Ie 4BL2+4CM2=5BC2 (6)AC2=122=144-----(i) AB2+BC2= (6√3)2+62 =108+36 AB2+BC2= 144---------(ii) From (i)&(ii) AC2=AB2+BC2(converse of Pythagoras theorem) ∠B=900 (7) In ∆PMR PR2= PM2+ MR2 = 62+82 = 36+64

A

B C

D

B

A C

M

L

P a g e 39 | 118

= 100 PR= 10cm In ∆PQR PQ2= QR2-PR2 = 262- 102 =676-100 =576 PQ= 24cm Now Area of ∆PQR= ½ x PR x PQ = ½ x 10 x24 = 120 cm2

8. Ratio of areas of two similar ∆s is equal to the ratio of squares of corresponding sides So Ratio of areas of two similar ∆s =(2x/3x)2 = 4/9 So Ratio of areas of two similar ∆s = ratio of squares of their corresponding altitudes= 4/9 So, Ratio of corresponding altitudes = 4/9

9. P divide CA in the ratio 2 :1 Therefore CP = 2/3 AC ………………………………..(i) QC = 2/3BC …………………………………(ii) In Right Triangle ACQ,

AQ2 = QC2 + AC2 Or, AQ2 = 4/9 BC2 + AC2 (QC = 2/3 BC) Or, 9 AQ2 = 4 BC2 + 9 AC2 ………………(iii) Similarly, In Right Triangle BCP 9BP2= 9BC2 + 4 AC2 ………………………(iv) Adding eq. (iii) & (iv) 9(AQ2 + BQ2) = 13(BC2 + AC2) 9(AQ2 + BQ2) =13AB2 10.In triangle ABD, PQ !! AB PQ/AB= DQ/BD Or, Z/X=DQ/BD……………………………..(i) In triangle BCD, PQ !! CD PQ/CD=BQ/BD Or, Z/Y=BQ/BD……………………………(ii) Adding eq. (i) & (ii) Z/X+ Z/Y = DQ/BD + BQ /BD = DQ + BQ/BD Or, Z/X + Z/Y = BD/BD=1 Or, 1/X + 1/y = 1/Z 11. State and Prove Pythagoras Theorem AP = AB – PB = ( 18- 12 )m = 6m [ PB = CD = pm ] Pc = BD = 8m In ∆ACP AC = √AP2 + PC2

= √(8)2+ (6)2 = √64 + 36 = √100 = 10 AC = 10 m (12)DE//GF &AC cuts them

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⇨∠DAG= ∠FGC(corres. ∠s) ∠GDE=900⇨∠GDA=900 ∆ADG ~∆GCF (By AA corollary ,shown above) (ii) similarly∆FEB ~∆GCF Since ∆ADG &∆FEB are both similar to ∆GCF ⇨∆ADG~∆FEB (iii)∆ADG~∆FEB AD/FE=DG/FB ⇨AD/DG= EF/EB (iv) ∆ADG~∆FEB AD/FE=DG/FB ⇨AD/DE= DE/EB(FE=DG=DE) DE2=AD X EB (13)(i)distance from the initial point=√32+42 = √25 =5m (ii) Pythagoras theorem (iii) To save time &energy

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INTRODUCTION TO TRIGONOMETRY

IMPORTANT CONCEPTS (TAKE A LOOK) 1. TRIGONOMETRY---A branch of mathematics in which we study the relationships between the sides and angles of a triangle, is called trigonometry.

2. TRIGONOMETRIC RATIOS -----Trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and length of its sides.

Trigonometric ratios of an acute angle in a right angled triangle ---

For ∠ β, sinβ= AB/ AC, cosβ = BC/AC, tanβ= AB/BC Cosecβ = AC/AB, secβ= AC/BC, cotβ = BC/AB

3. Relationship between different trigonometric ratios

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4. Trigonometric Identity---- An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angle. Important trigonometric identities:

(i) sin2 + cos2 =1

(ii) 1 + tan2 = sec2

(iii) 1 +cot2 = cosec2 5. Trigonometric Ratios of some specific angles.

0o 30o 45o 60o 90o

sin 0 ½ 1/2 3/2 1

cos 1 3/2 1/2 1/2 0

tan 0 1/3 1 3 Not defined

cot Not defined

3 1 1/3 0

sec 1 2/3 2 2 Not defined

cosec Not defined

2 2 2/3 1

6. Trigonometric ratios of complementary angles.

(i) sin (90o - ) = cos

(ii) cos (90o - ) = sin

(iii) tan (90o - ) = cot

(iv) cot (90o - ) = tan

(v) sec (90o - ) = cosec

(vi) cosec (90o - ) = sec

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Level – I

1. If θ and 3θ-30° are acute angles such that sinθ=cos (3θ-30°), then find the value of tanθ.

2. Find the value of

3. Find the value of (sinθ+cosθ) ²+ (cosθ-sinθ) ²

4. If tanθ= then find the value of cos²θ-sin²θ

5. If secθ+tanθ=p, then find the value of secθ-tanθ

6. Change sec⁴θ-sec²θ in terms of tanθ.

7. Prove that sin3α +cos3α

+ Sinα cosα= 1 (CBSE 2009)

Sinα +cosα

8. In a triangle ABC, it is given that < C= 90˚ and tanA=1/√3, find the value of (sinA cosB +cosA sinB)

(CBSE 2008)

9. Find the value of cosec267˚- tan223˚.

10. Ifcos x=cos60° cos30°+sin60° sin30°, then find the value of x

11. If 0°≤ x ≤90° and 2sin²x=1/2, then find the value of x

12. Find the value of cosec²30°-sin²45°-sec²60°

13. Simplify (secθ+tanθ) (1-sinθ)

14. Prove that cosA/ (1-sinA) +cosA/ (1+sinA) =2secA

Level – II

1. If secα=5/4 then evaluate tanα/ (1+tan²α).

2. If A+B =90°, then prove that√𝑡𝑎𝑛𝐴.𝑡𝑎𝑛𝐵+𝑡𝑎𝑛𝐴.𝑐𝑜𝑡𝐵𝑠𝑖𝑛𝐴.𝑠𝑒𝑐𝐵

−𝑠𝑖𝑛2𝐵

𝑐𝑜𝑠2𝐴 = tanA

3. If 7 sin2A +3 cos2A= 4, show that tanA =1/√3. (CBSE 2008)

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4. Prove that . + = 2cosecA

5. Prove that (sinθ+cosecθ) ² + (cosθ+secθ) ² =7+tan²θ+cot²θ. (CBSE 2008, 2009C)

6. Evalute --

7. Find the value of sin30˚ geometrically.

8. If tan (A-B) =√3, and sinA =1, then find A and B.

9. If θ is an acute angle and sinθ=cosθ, find the value of 3tan²θ + 2sin²θ – 1.

10. If cosθ + sin θ = 1 and sinθ – cosθ = 1, prove that x²/a² + y²/b² = 2.

11. Provethat = tanθ.

Level - III

1. Evaluate the following: - sin²25° + sin²65° + (tan5° tan15° tan30° tan75° tan85°)

2. If = m, and = n, show that (m²+n²) cos²β = n². (CBSE 2012)

3. Prove that tan²θ + cot²θ + 2 = cosec²θ sec²θ

4. If cosθ + sinθ = √2 cos θ, then show that (cosθ-sinθ) = √2 sinθ.

(CBSE 1997, 2002, 2007)

5. Prove that (sinθ+secθ) ² + (cosθ + cosecθ) ² = (1+secθ cosecθ) ².

6. Prove that sinθ/ (1-cosθ) + tanθ/ (1+cosθ) = secθcosecθ + cotθ.

7. If x = asinθ and y = btanθ. Prove that a2/x2 – b2/y2 = 1.

8. Prove that sin 6θ + cos6θ = 1- 3sin2θcos2θ.

9. Prove that (secθ+tanθ – 1)/ (tanθ – secθ+1) = cosθ/ (1 – sinθ).

P a g e 45 | 118

10. Prove that (1 +cotθ - cosec θ) (1+tanθ+secθ) = 2 (CBSE 2005, 07, 08)

11.Evaluate -

12. If sinθ +cosθ =m and secθ+ cosecθ =n, then prove that n (m2 – 1) =2m.

Self-Evaluation

1. If a cosθ + b sinθ = c, then prove that asinθ – bcosθ = ∓

2. If A,B,C are interior angles of triangle ABC, show that cosec²( ) - tan² = 1.

3. Ifsinθ + sin²θ + sin³θ = 1, prove that cos⁶θ – 4cos⁴θ + 8cos²θ = 4.

4. IftanA = ntanB, sinA = msinB, prove that cos²A = (m² - 1)/(n²-1).

5. Evaluate: secθcosec (90°- θ) – tanθ cot (90° - θ) + sin²55°+ sin²35°

tan10˚tan20˚tan60˚tan70˚tan80˚

6. If secθ + tanθ=p, prove that sinθ = (p²-1)/ (p²+1).

7. Prove that - = - .

8. Prove that: + = sinθ + cosθ

9. Prove that = .

10. Prove that(1 + cosθ + sinθ) / (1+ cosθ – sinθ) = (1 + sinθ)/ cosθ

P a g e 46 | 118

STATICTICS

(i) Assumed Mean method or Shortcut method

Mean = = a +

Where a = assumed mean And di= Xi - a

(ii) Step deviation method.

Mean = = a +

Where a = assumed mean h = class size And ui= (Xi – a)/h

Median of a grouped frequency distribution can be calculated by

Median = l +

Where l = lower limit of median class n = number of observations cf = cumulative frequency of class preceding the median class f = frequency of median class h = class size of the median class.

Mode of grouped data can be calculated by the following formula.

Mode = l +

Where l = lower limit of modal class h = size of class interval f1 = Frequency of the modal class fo = frequency of class preceding the modal class f2= frequency of class succeeding the modal class

Empirical relationship between the three measures of central tendency. 3 Median = Mode + 2 Mean Or, Mode = 3 Median – 2 Mean

Ogive Ogive is the graphical representation of the cumulative frequency distribution. It is of two types: (i) Less than type ogive. (ii) More than type ogive

P a g e 47 | 118

Median by graphical method The x-coordinated of the point of intersection of ‘less than ogive’ and ‘more than ogive’ gives the median.

LEVEL – I

Slno Question

1 What is the mean of 1st ten prime numbers?

2 What measure of central tendency is represented by the abscissa of the point where less than ogive and more than ogive intersect?

3 If the mode of a data is 45 and mean is 27, then median is ___________.

4 Find the mode of the following

Xi 35 38 40 42 44

fi 5 9 10 7 2

5 Write the median class of the following distribution.

Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70

Frequency 4 4 8 10 12 8 4

6 The wickets taken by a bowler in 10 cricket matches are as follows: 2, 6 ,4 ,5, 0, 2, 1, 3, 2, 3 Find the mode of the data

7. How one can find median of a frequency distribution graphically

8. What important information one can get by the abscissa of the point of intersection of the less than type and the more than type commulative frequency curve of a group data

LEVEL – II

Slno Question Ans

1 Find the median of the following frequency distribution

Height in cm 160-162 163-165 166-168 169-171 172-174

Frequency 15 117 136 118 14

167

2 Given below is the distribution of IQ of the 100 students. Find the median IQ

IQ 75-84 85-94 95-104 105-114

115-124 125-134 135-144

Frequency 8 11 26 31 18 4 2

106.1

3 Find the median of the following distribution

Class interval 0-10 10-20 20-30 30-40 40-50 50-60

Frequency 5 8 20 15 7 5

28.5

4 A class teacher has the following absentee record of 40 students of a class for the whole

P a g e 48 | 118

term.

No. of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40

No. of students

11 10 7 4 4 3 1

Write the above distribution as less than type cumulative frequency distribution.

5

Using the assumed mean method find the mean of the following data.

Class interval 0-10 10-20 20-30 30-40 40-50

frequency 7 8 12 13 10

Ans 27.2

6 Name the keyword used for central tendency Mean , median , mode

LEVEL – III

SN Question Ans

1 If the mean distribution is 25

Class 0-10 10-20 20-30 30-40 40-50

Frequency 5 18 15 P 6

Then find p.

P=16

2 Find the mean of the following frequency distribution using step deviation method

Class 0-10 10-20 20-30 30-40 40-50

Frequency 7 12 13 10 8

25

3 Find the value of p if the median of the following frequency distribution is 50

Class 20-30 30-40 40-50 50-60 60-70 70-80 80-90

Frequency 25 15 P 6 24 12 8

P=10

4 Find the median of the following data

Marks Less Than

10

Less Than

30

Less Than

50

Less Than

70

Less Than 90

Less Than 110

Less Than 130

Less than 150

Frequency 0 10 25 43 65 87 96 100

.

76.36

5 Compare the modal ages of two groups of students appearing for entrance examination.

Age in yrs 16-18 18-20 20-22 22-24 24-26

Group A 50 78 46 28 23

P a g e 49 | 118

Group B 54 89 40 25 17

6 The mean of the following frequency distribution is 57.6 and the sum of the observations is 50. Find the missing frequencies f1 and f2.

Class 0-20 20-40 40-60 60-80 80-100 100-120 Total

Frequency 7 f1 12 f2 8 5 50

f1 =8 and f2 =10

7 The following distribution give the daily income of 65 workers of a factory

Daily income (in

Rs)

100-120 120-140 140-160 160-180 180-200

No. of workers

14 16 10 16 9

Convert the above to a more than type cumulative frequency distribution and draw its ogive.

8 Draw a less than type and more than type ogives for the following distribution on the same graph. Also find the median from the graph.

Marks 30-39 40-49 50-59 60-69 70-79 80-89 90-99

No. of students

14 6 10 20 30 8 12

SELF – EVALUATION

1. What is the value of the median of the data using the graph in figure of less than ogive and more than ogive?

2. If mean =60 and median =50, then find mode using empirical relationship. 3. Find the value of p, if the mean of the following distribution is 18.

Variate (xi) 13 15 17 19 20+p 23

Frequency (fi)

8 2 3 4 5p 6

4. Find the mean, mode and median for the following data.

Classes 0-10 10-20 20-30 30-40 40-50 50-60 60-70

frequency 5 8 15 20 14 8 5

P a g e 50 | 118

5. The median of the following data is 52.5. find the value of x and y, if the total frequency is 100.

Class Interval

0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

frequency 2 5 X 12 17 20 Y 9 7 4

6. Draw ‘less than ogive’ and ‘more than ogive’ for the following distribution and hence find its median.

Classes 20-30 30-40 40-50 50-60 60-70 70-80 80-90

frequency 10 8 12 24 6 25 15

7. Find the mean marks for the following data.

Marks Below 10

Below 20

Below 30

Below 40

Below 50

Below 60

Below 70

Below 80

Below 90

Below 100

No. of students

5 9 17 29 45 60 70 78 83 85

8. The following table shows age distribution of persons in a particular region. Calculate the median age.

Age in years

Below 10

Below 20

Below 30

Below 40

Below 50

Below 60

Below 70

Below 80

No. of persons

200 500 900 1200 1400 1500 1550 1560

9. If the median of the following data is 32.5. Find the value of x and y.

Class Interval

0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total

frequency x 5 9 12 y 3 2 40

10. The following are ages of 300 patients getting medical treatment in a hospital on a particular day.

Age( in

years)

10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

Number of

patients

60 42 55 70 53 20

Draw:

1. Less than type cumulative frequency distribution

2. More than type cumulative frequency distribution

P a g e 51 | 118

Value Based Question

Q1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a

locality.

Monthly

consumption

(in units)

65 – 85 85 – 105 105 – 125 125- 145 145- 165 165 – 185 185 – 205

Number of

consumers

4 5 13 20 14 8 4

Mr. Sharma always saves electricity by switching of all the electrical equipment just immediately after their uses. So , his family belongs to the group 65- 85 .

(i) Find the median of the above data (ii) How many families consumed 125 or more units of electricity during a month? (iii) What moral values of Mr. Sharma have been depicted in this situation?

Q2. The mileage (km per litre) of 50 cars of the same models is tested by manufacturers and details are

tabulated as given below:-

Mileage (km per

litre) 10 – 12 12 – 14 14 - 16 16- 18

No. of cars 7 12

18 13

i. Find the mean mileage.

ii. The manufacturer claims that the mileage of the model is 16km/litre. Do you agree with this

claim?

iii. Which values do you think the manufacturer should imbibe in his life?

P a g e 52 | 118

ANSWER 1. 12.9

2. MEDIAN

3. 33

4. MODE = 40

5. MEDIAN = 30-40

6. 2

7. OGIVE

8. Median

Level II Q1 167

Q2 106.1

Q3 28.5l

Q4

No. of days Less

Than 6

Less

Than 10

Less

Than 14

Less

Than 20

Less

Than 28

Less

Than 38

Less

Than 40

No. of

students

11 21 28 32 36 39 40

Q5 27.2

Q6 Mean, median, mode

P a g e 53 | 118

MODEL SAMPLE PAPER – SA 1

BLUE PRINT CLASS-X Sub:-Mathematics

S. N. Name of Chapter

VSA SA-I SA-II LA Total

1 Number system

2(2) 1(2) 1(3) 1(4) 5(11)

2 Algebra 1(1) 2(4) 2(6) 3(12) 8(23)

3 Geometry 1(1) 1(2) 2(6) 2(8) 6(17)

4 Trigonometry ------ 1(2) 4(12) 2(8) 7(22)

5 Statistics ------ 1(2) 1(3) 3(12) 5(17)

Total 4(4) 6(12) 10(30) 11(44) 31(90)

Note: - Number of questions is given outside the brackets and marks are given within the bracket.

P a g e 54 | 118

MODEL SAMPLE PAPER – SA1 Time Allowed: - 3 hours Max. Marks:-90

General instruction:- (i). Question should be distributed to the students before 15 minutes of the commencement of examination.

(ii). All questions are compulsory.

(iii). The questions paper comprises of 31 questions divided into four sections A, B, C and D. you are to attempt all

the four sections.

(iv). Question no. 1 to 4 in section ‘A’ is of 1 mark each.

Question no. 5 to 10 in section ‘B’ are of 2 marks each. Question no. 11 to 20 in section ‘C’ are of 3 marks each. Question no. 21 to 31 in section ‘D’ are of 4 marks each.

(v). Use of calculator is not permitted.

lkekU; funsZ”k

(i). iz”u i= Nk=ksa dks ijh{kk “kq: gksus ds 15 feuV igys forfjr djuk gSA

(ii). lHkh iz”u vfuok;Z gSSA

(iii). bl iz”u i= esa dqy 31 iz”u gS ftUgs pkj [k.Mksa esa ckaVk x;k gSA

(iv). [kaM esa 1 ls 4 iz”u gSA izR;sad iz”u 1 vad dk gSA [kaM esa 5 ls 10 iz”u gSA izR;sad iz”u 2 vad dk gSA

[kaM esa 11 ls 20 iz”u gSA izR;sad iz”u 3 vad dk gSA

[kaM esa 21 ls 31 iz”u gSA izR;sad iz”u 4 vad dk gSA

(v). dsYdqysVj dk mi;ksax oftZr gSA

Section: - A 1. If the HCF of 55 and 99 is expressible in the form of 55m - 99 then find the value of m.

;fn 55 vkSj 99 dk HCF dks 55m&99 ds :Ik O;Dr fd;k x;k gks rks m dk eku fudkysA 2. If

241

4000=

241

2𝑚𝑋 5𝑛, find the values of m and n where m & n are whole number.

;fn241

4000=

241

2𝑚𝑋 5𝑛, rks m vkSj n dk eku Kkr djsa tgkW m vkSj n iw.kZ la[;k,W gSA

3. For what value of K, (-4) is a zero of the polynomial 𝑥2 − 𝑥 − (2𝑘 + 2)?

“K” ds fdl eku ds fy, &4] cgqin 𝑥2 − 𝑥 − (2𝑘 + 2) dk “kqU;kad gksxk\ 4. In ∆𝐴𝐵𝐶 shown in figure DE||BC. If BC =8cm, DE=6cm and area of ∆ADE=45cm2, what is the area of ∆𝐴𝐵𝐶.

A

D E

B C fn;s x;s fp= ds vuqlkj ∆𝐴𝐵𝐶 es DE||BC. ;fn BC =8cm, DE=6cm vkSj ∆ADE dk {kas+=Qy =45cm2 rks ∆𝐴𝐵𝐶. dk {kas=Qy D;k gksxk\

Section:-B 5. Find a quadratic polynomial whose zero are -2 and 3.

,d f}?kkr cgqin Kkr djsa ftldk “kqU;kad &2 vkSj 3 gSA

6. Check whether 6n can end with the digit zero for any natural number n.

tkWp djsa fd vad 6n ”kwU; ds lkFk lekIr gks ldrk gS ;k ugha] tgkW n ,d izkd`r la[;k gSA 7. The larger of two supplementary angles exceeds smaller by 200. Find the angles.

nks laiwjd dks.kksa esa ls lcls cM+k dks.k] NksVs dks.k ls 200 cM+k gSA rks dks.kksa dks Kkr djsaA 8. In the given fig DE||BC, if BD=x – 3, AB=2x, CE=x – 2 and AC=2x+3. Find x.

A

P a g e 55 | 118

D E

B C fn;s x;s fp= esa DE||BC] ;fn BD=x – 3, AB=2x, CE=x – 2 vkSj AC=2x+3. Rkks X dk eku fudkysA

9. If cos𝜃=𝑥

𝑦 then find the value of tan 𝜃 & sec𝜃.

;fn cos𝜃=𝑥

𝑦 rks tan 𝜃 vkSj sec𝜃 dk eku fudkysA

10. If the mean of the following data is 15, find P.

X 5 10 15 20 25

F 6 P 6 10 5

fuEufyf[kr lkj.kh dk ek/; ;fn 15 gS rks P dk eku fudkysA X 5 10 15 20 25

F 6 P 6 10 5

Section: - C

11. Prove that √7 is irrational number.

fl) djs fd √7 ,d vifjes; la[;k gSA 12. For what value of P will the following pair of linear equations have infinitely many solutions?

(P-3) x + 3y = p; p x + p y = 12

fuEufyf[kr ;qXe jSf[kd lfedj.kksa esa P ds fdl eku ds fy, vuar gy gksxk\ (P-3) x + 3y = p; p x + p y = 12

13. Solve for x & y: 𝑥+1

2+

𝑦−1

3= 8;

𝑥−1

3+

𝑦+1

2= 9

X vkSj y dk gy fudkysA 𝑥+1

2+

𝑦−1

3= 8;

𝑥−1

3+

𝑦+1

2= 9

14. D and E are points on the side CA and CB respectively of ∆𝐴𝐵𝐶 right angled at C. prove that

AE2 + BD2 = AB2 + DE2

Ledks.k f=Hkqt ABC tks C ij ledks.k gSA D vkSj E dze”k% Hkqtk, CA vkSj CB ds fcUnq gS rks fl) djs fd AE2 + BD2 = AB2 + DE2.

15. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding

medians.

fl) djs fd nks ledks.k f=Hkqtks ds {ks=Qy dk vuqikr mlds rnuq:ih ekf/;dkvksa ds oxZ ds vuqikr ds cjkcj

gksrh gSA

16. Evaluate:- 5𝑠𝑖𝑛2300 + 𝑐𝑜𝑠2450+4𝑡𝑎𝑛2600

2𝑠𝑖𝑛300.𝑐𝑜𝑠600+𝑡𝑎𝑛450

Ekku fudkysA

5𝑠𝑖𝑛2300 + 𝑐𝑜𝑠2450+4𝑡𝑎𝑛2600

2𝑠𝑖𝑛300.𝑐𝑜𝑠600+𝑡𝑎𝑛450

17. If sin (A + B) =1 and cos (A – B) =√3

2, find A and B.

;fn sin (A + B) =1 vkSj cos (A – B) =√3

2, rks A avkSj B dk eku fudkysaA

18. In the figure, ∆𝐴𝐵𝐶is a right angled triangle, D is the mid-point of BC.

P a g e 56 | 118

fl) djs fd (1 + cot A –cosec A) ( 1+ tan A + sec A) = 2. 20. The mean of the following frequently table is 50. Find the value of x & y.

Class interval

0-20 20-40 40-60 60-80 80-100 Total

Frequency 17 X 32 Y 19 120

fuEufyf[kr caVu lkj.kh es ek?; 50 gS] rks X vkSj Y dk eku Kkr djsaA Class

interval 0-20 20-40 40-60 60-80 80-100 Total

Frequency 17 X 32 Y 19 120

Section: - D 21. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m +1 or 9m + 8.

;qfDyM fMfotu ysaEek dk iz;ksx djrs gq, fn[kk,W fd fdlh Hkh /kukRed la[;kvksa dk ?ku 9m, 9m +1 vFkok 9m + 8 ds :Ik es gksrk gSA

22. It two zeroes of the polynomial 𝑥4 − 6𝑥3 − 26𝑥2 + 138𝑥 − 35 𝑎𝑟𝑒 2 + √3 & 2 − √3. Find other Zeros.

;fn cgqin 𝑥4 − 6𝑥3 − 26𝑥2 + 138𝑥 − 35 ds nks “kwU;kad 2 + √3 & 2 − √3 gSA rks nwljs “kwU;kadksa dks Kkr djsaA

23. Solve for x & y : 1

3𝑥+𝑦+

1

3𝑥−𝑦=

3

4

1

2(3𝑥+𝑦)−

1

2(3𝑥−𝑦)=

−1

8

x vkSj y dk gy fudkysA 1

3𝑥+𝑦+

1

3𝑥−𝑦=

3

4

1

2(3𝑥+𝑦)−

1

2(3𝑥−𝑦)=

−1

8

24. A boat goes 30 km upstream and 44 km downstream in 10 hrs. In 13 hours, it can go 40 km upstream and 55 km

downstream. Determine the speed of the stream and that of the boat in still water.

,d uko 10 ?kaVsa es 30 fdyksehVj ?kkjk ds foijhr vkSj 44 fd0 eh0 /kkjk dh fn”kk es tkrh gSA] 13 ?kaVsa es 40

fd0 eh0 ?kkjk ds foijhr vkSj 55 fd0 eh0 /kkjk dh fn”kk esa tk ldrh gSA fLFkj ty esa uko vkSj /kkjk dk osx

Kkr djsaA

25. State and prove Pythagoras theorem.

ikbFkkxksjl izes; dks fy[ksa vkSj fl) djsaA

26. In an equilateral triangle ABC, D is a point on side BC such that BD= 1

3𝐵𝐶. 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 9𝐴𝐷2 = 7𝐴𝐵2

fdlh leckgq f=Hkqt ABC esa D, BC ij ,d fcUnq bl rjg gS fd BD= 1

3𝐵𝐶. fl) djsa fd 9𝐴𝐷2 = 7𝐴𝐵2

27. If Cos Ɵ – Sin Ɵ = √2𝑆𝑖𝑛Ɵ, prove that Cos Ɵ + Sin Ɵ = √2Cos Ɵ.

;fn Cos Ɵ – Sin Ɵ = √2𝑆𝑖𝑛Ɵ, rks fl) djsa fd Cos Ɵ + Sin Ɵ = √2Cos Ɵ.

28. Prove that

𝑡𝑎𝑛Ɵ

1−𝐶𝑜𝑡Ɵ+

𝐶𝑜𝑡Ɵ

1−𝑡𝑎𝑛Ɵ= 1 + 𝑠𝑒𝑐Ɵ. 𝑐𝑜𝑠𝑒𝑐Ɵ

fl) djsa fdA 𝑡𝑎𝑛Ɵ

1−𝐶𝑜𝑡Ɵ+

𝐶𝑜𝑡Ɵ

1−𝑡𝑎𝑛Ɵ= 1 + 𝑠𝑒𝑐Ɵ. 𝑐𝑜𝑠𝑒𝑐Ɵ

29. Calculate the arithmetic mean of the following frequency distribution using the step deviation method.

Class Interval

0-50 50-100 100-150 150-200 200-250 250-300

Frequency 17 35 43 40 21 24

fuEufyf[kr caVu lkj.kh es ek?; in fopyu fof/k ls Kkr djsaA

Class Interval

0-50 50-100 100-150 150-200 200-250 250-300

P a g e 57 | 118

Frequency 17 35 43 40 21 24

30. To highlight child Labour problem, some students organized a javelin through competition. 50 students

participated in this completion. The distance (in meters) thrown are recorded below.

Distain (in m) 0-20 20-40 40-60 60-80 80-100

Number of students

6 11 17 12 04

a. Construct a cumulative frequency table.

b. Draw cumulative frequencies curve (Less than type) and calculate the median distance thrown.

c. Which value is depicted by students?

Ckky etnqjh dh leL;k dks mtkxj djus ds fy,] dqN fo|kFkhZ tkoyhu Fkzks izfr;ksfxrk ds fy, laxfBr

gq,A bl izfr;ksfxrk esa 50 fo|kFkhZ;ksa us Hkkx fy;kA

Qsadh x;h nwfj;ksa dk fjdkMZ fuEufyf[kr gSA

nwjh (feVj esa) 0-20 20-40 40-60 60-80 80-100

fo|kFkhZ;ksa dh la[;k 6 11 17 12 04

(i). mij fyf[kr lkj.kh dh lgk;rk ls lap;h ckjackjrk lkj.kh cukosaA

(ii). laap;h ckjackjrk odz [khpsa vkSj Qasdh x;h nwfj;ksa dh ekf/;dk dh x.kuk djsaA

(iii). fo|kFkhZ }kjk dkSu &lk ewY; n”kkZrk gS \ 31. Compare the modal a ages of two groups of students A and B appearing for an entrance test.

Age (in Year) Group:-A Group:-B

16-18 50 54

18-20 78 89

20-22 46 40

22-24 28 25

24-26 23 17

fo|kFkhZ ds nks lewg A vkSj B tks ,d izfr;ksfxrk ijh{kk es lfEefyr gksrs gS] ds cgqyd mez dh rqyuk djsa A

mez ¼o’kZ es½ Lkeqg %& A Lkeqg %&B 16-18 50 54

18-20 78 89 20-22 46 40

22-24 28 25 24-26 23 17

P a g e 58 | 118

MODEL SAMPLE PAPER MARKING SCHEME

CLASS –X SA-1

1. m = 2 1

2. m= 5, n=3 1

3. K=9 1

4. ar∆𝐴𝐵𝐶 = 80𝑐𝑚2 1

Section-B 5. p(x)= 𝑥2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽 1

=𝑥2 − (−2 + 3)𝑥 + (−6) 1

2

𝑝(𝑥) = 𝑥2 − 𝑥 − 6 1

2

6. 6𝑛 = 2𝑛 𝑋 3𝑛 1

∴ 6𝑛 ℎ𝑎𝑠 𝑛𝑜 5 𝑎𝑠 𝑓𝑎𝑐𝑡𝑜𝑟. 1

2

∴ 6𝑛 𝑐𝑎𝑛 𝑛𝑜𝑡 𝑒𝑛𝑑 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑑𝑖𝑔𝑖𝑡 𝑧𝑒𝑟𝑜. 1

2

7. Let smaller angle =𝑥0

∴ Larger angle = (𝑥 + 20)0 1

2

X+x+20 = 1800 ⇒X=800 1 ∴ Smaller angle = 800

Larger angle = 1000 1

2

8. ∵DE||BC

⇒𝐵𝐷

𝐴𝐵=

𝐶𝐸

𝐴𝐶

1

2

⇒𝑥−3

2𝑥=

𝑥−2

2𝑥+3 1

⇒X = 9 1

2

9. 𝑐𝑜𝑠𝜃 = 𝑥

𝑦

𝑡𝑎𝑛𝜃 =𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃=

√1−𝑐𝑜𝑠2𝜃

𝑐𝑜𝑠𝜃 1

=√1−

𝑥2

𝑦2

𝑥

𝑦

= √𝑦2−𝑥2

𝑥

1

2

𝑠𝑒𝑐𝜃 =1

𝑐𝑜𝑠𝜃=

𝑦

𝑥

1

2

10.

X 𝑓 𝑓(𝑥) 5 6 30

10 P 10p

15 6 90

20 10 200

25 5 125

27+p 445+10p

X=445+10𝑝

27+𝑝 1

15 = 445+10𝑝

27+𝑝

1

2

⇒P=8 1

2

11. Let √7 is a rational no.

P a g e 59 | 118

∴ √7 =𝑝

𝑞

(𝑝 & 𝑞 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠, 𝑞 ≠ 0, ℎ𝑎𝑣𝑖𝑛𝑔 𝑛𝑜 𝑐𝑜𝑚𝑚𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟) 1

2

Squaring

7 = 𝑝2

𝑞2

⇒𝑝2 = 7𝑞2--------- (i) 1 ∴ 𝑝2 & 𝑝 𝑎𝑟𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 7 𝑙𝑒𝑡 𝑝 = 7𝑚 𝑝2 = 49𝑚2----------- (ii) 𝑓𝑟𝑜𝑚 (𝑖)𝑎𝑛𝑑 (𝑖𝑖) 𝑞2 = 7𝑚2

⇒𝑞2& 𝑞 𝑎𝑟𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 7 ∴ 𝑝& 𝑞 ℎ𝑎𝑣𝑒 7 𝑎𝑠 𝑐𝑜𝑚𝑚𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟. 1 ∴ 𝑜𝑢𝑟 𝑠𝑢𝑝𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑖𝑠 𝑤𝑟𝑜𝑛𝑔

∴ √7𝑖𝑠 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟. 1

2

12. a1=p-3, b1=3, c1=p

a2=p, b2=p, c2=12 1

2

For infinitely many solution

𝑎1

𝑎2=

𝑏1

𝑏2=

𝑐1

𝑐2 1

⇒𝑝−3

𝑝=

3

𝑝=

𝑝

12

From the 1st and 2nd 1 p=0, p=6 from last two p=6, -6

⇒p=6 Ans. 1

2

13. 𝑥+1

2+

𝑦−1

3= 8

⇒3x + 2y =47------- (i) 1 𝑥−1

3+

𝑦+1

2= 9

⇒2x+3y=53-------- (ii) 1 Solving (i) & (ii) X=7, y=13 1

14. A

D

C E B In right ∆𝐴𝐸𝐶, 𝐴𝐸2 = 𝐴𝐶2 + 𝐶𝐸2--------------------- (i) 1 In right ∆𝐷𝐵𝐶 𝐵𝐷2 = 𝐶𝐷2 + 𝐵𝐶2 ------------------ (ii) Adding (i) & (ii) 𝐴𝐸2 + 𝐵𝐷2 = 𝐴𝐶2 + 𝐶𝐸2 + 𝐶𝐷2 + 𝐵𝐶2 1

= 𝐴𝐵2 + 𝐷𝐸2 Proved 1

2

15. A P

B D C Q M R

P a g e 60 | 118

∆𝐴𝐵𝐶 ≈ ∆𝑃𝑄𝑅 1

2

⇒ 𝐴𝐵

𝑃𝑄=

𝐵𝐶

𝑄𝑅=

𝐴𝐶

𝑃𝑅 1

⇒ 𝐴𝐵

𝑃𝑄=

𝐵𝐷

𝑄𝑀=

𝐵𝐷

𝑄𝑀

In ∆𝐴𝐵𝐶 & ∆𝑃𝑄𝑀 𝐴𝐵

𝑃𝑄=

𝐵𝐷

𝑄𝑀 & < 𝐵 = < 𝑄 (𝑔𝑖𝑣𝑒𝑛)

⇒∆𝐴𝐵𝐷 ≈ ∆𝑃𝑄𝑀

⇒ 𝐴𝐵

𝑃𝑄=

𝐴𝐷

𝑃𝑀

⇒ 𝑎𝑟∆𝐴𝐵𝐶

𝑎𝑟∆𝑃𝑄𝑅=

𝐴𝐷2

𝑃𝑀2

1

2

16. 5 𝑋 𝑠𝑖𝑛2300 + 𝑐𝑜𝑠2450 + 4𝑡𝑎𝑛2600

2𝑠𝑖𝑛300.𝑐𝑜𝑠600+ 𝑡𝑎𝑛450

=5 𝑋 (

1

2)2+ (

1

√2)2+4 𝑋 (√3)

2

2 𝑋 1

2 𝑋

1

2 + 1

2

=5

4+

1

2+12

1

2+1

=55

43

2

1

2

=55

6𝐴𝑛𝑠

1

2

17. Sin( A + B ) = Sin 900

⇒A+B =900---------- (i) 1 Cos (A - B) = 𝑐𝑜𝑠600 ⇒ A-B =60 1 ⇒ A = 75 & B =15 1

18.

D is the midpoint of BC A ∴BD =DC 1

𝑡𝑎𝑛𝜃

𝑡𝑎𝑛∅=

𝐴𝐶

𝐵𝐶𝐴𝐶

𝐷𝐶

= 𝐷𝐶

𝐵𝐶 Ɵ ∅ 1

= 𝐷𝐶

2𝐷𝐶=

1

2 B D C 1

19. LHS = (1 + Cot A – cosec A) (1 + tan A + sec A)

= (1 +cos 𝐴

𝑠𝑖𝑛 𝐴−

1

sin 𝐴) (1 +

sin 𝐴

cos 𝐴+

1

cos 𝐴) 1

= (sin 𝐴+cos 𝐴−1

sin 𝐴) (

cos 𝐴+sin 𝐴+1

cos 𝐴)

= (sin 𝐴+cos 𝐴)2− (1)2

sin 𝐴.cos 𝐴 1

=𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴+2𝑐𝑜𝑠 𝐴.𝑠𝑖𝑛𝐴−1

𝑠𝑖𝑛𝐴.cos 𝐴

1

2

=1+2 cos 𝐴.𝑠𝑖𝑛𝐴−1

𝑠𝑖𝑛𝐴.𝑐𝑜𝑠𝐴

= 2 = 𝑅𝐻𝑆. 1

2

20.

C.I Fi Xi ui Fiui

0-20 17 10 -2 -34

20-40 X 30 -1 -x

40-60 32 50 = 9 0 0

60-80 Y 70 1 Y

80-100 19 90 2 38

∑𝑓𝑖 = 120 ∑𝑓𝑖𝑢𝑖=4−𝑥+𝑦 68 + x + y = 120 X + y = 52-------------------------- (i)

P a g e 61 | 118

X= a + h ∑𝑓𝑖𝑢𝑖∑𝑓𝑖

1

2

50 = 50 = 20 x4−𝑥+𝑦

120 ………………………..(ii)

1

2

Solving (i) & (ii) X = 28

Y = 24 1

2

21. Let a = 3q +r, 0 ≤ r < 3

⇒r = 0, 1, 2 Let r = 0 ∴ a = 3q 𝑐𝑢𝑏𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒 𝑎3 = 27𝑞3 9 𝑋 (3𝑞3) = 9 𝑋 𝑚. 1 Let. r =1 a = 3q +1 𝑎3 = (3𝑞 + 1)3 (Cubing) 𝑎3 = 27𝑞3 + 3 𝑋 (3𝑞)2 𝑋 1 + 3 𝑋 3𝑞 𝑋 1 + 13 = 27𝑞3 + 27𝑞2 + 9𝑞 + 1 = 9(3𝑞3 + 3𝑞2 + 𝑞 ) + 1 = 9𝑚 + 1 1 Let. r =2 a = 3q +2 𝑎3 = (3𝑞 + 2)3 (Cubing)

𝑎3 = 27𝑞3 + 3 𝑋 (3𝑞)2 𝑋 2 + 3 𝑋 3𝑞 𝑋 22 + 23 1

2

= 27𝑞3 + 54𝑞2 + 36𝑞 + 8 = 9(3𝑞3 + 6𝑞2 + 4𝑞 ) + 8 = 9𝑚 + 8

Cube of any positive integer is of the form 1

2 qm, qm+1 or qm+8

22. P(x) = 𝑥4 − 6𝑥3 − 26𝑥2 + 138𝑥 − 35

( 𝑥 − 2 − √3)(𝑥 − 2 + √3) = 𝑥2 − 4𝑥 + 1 is a factor of p(x) 1

Now 𝑥4−6𝑥3−26𝑥2+138𝑥−35

𝑥2−4𝑥+1= 𝑥2 − 2𝑥 − 35 2

𝑥2 − 2𝑥 − 35 = 𝑥2 − 7𝑥 − 5𝑥 − 35 =(x-7)(x-5) ∴ Other zeroes are 7 & -5. 1

23. Let 1

3𝑥+𝑦= 𝑎,

1

3𝑥−𝑦= 𝑏

𝑎 + 𝑏 =3

4 ---------- (i) 1

𝑎

2−

𝑏

2= −

1

8

⇒𝑎 − 𝑏 =−1

4 ------------------ (ii) 1

From (i) & (ii)

𝑎 =1

4, 𝑏 =

1

2 1

⇒3x + y = 4 3x – y = 2 On solving x = 1, y=1 1

24. Let speed of boat in still water = X km/h.

Speed of stream = y Km/h.

Speed of boat in downstream = (x + y) km/h. Speed of boat in upstream =(x – y) km/h. 1 44

𝑥+𝑦 +

30

𝑥−𝑦= 10

P a g e 62 | 118

55

𝑥+𝑦+

40

𝑥−𝑦= 13 2

𝑜𝑛 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑥 = 8𝑘𝑚/ℎ, 𝑦 = 3𝑘𝑚/ℎ 1 25. Fig, given to prove, constructions. 2

Proof 2 26. A

B D E C

Const: - Draw AE⊥ 𝐵𝐶. 1

2

Proof:- ∠𝐵𝐸𝐶 = 900, 𝐵𝐸 = 𝐶𝐸(∴ 𝐴𝐸 𝑖𝑠 𝑚𝑒𝑑𝑖𝑎𝑛 𝑎𝑙𝑠𝑜).

𝐵𝐷 =1

3𝐵𝐶.

⊿𝐴𝐷𝐸, 1 𝐴𝐷2 = 𝐴𝐸2 + 𝐷𝐸2 = 𝐴𝐵2 − 𝐵𝐸2 + (𝐵𝐸 − 𝐵𝐷)2 = 𝐴𝐵2 + 𝐵𝐷2 − 2𝐵𝐸. 𝐵𝐷

= 𝐴𝐵2 (1

3𝐵𝐶)

2

− 𝐵𝐶 𝑋 𝐵𝐶

3

𝐴𝐵2 +𝐴𝐵2

9−

𝐴𝐵2

3 [∵ 𝐴𝐵 = 𝐵𝐶 = 𝐴𝑐] 1

⇒ 9𝐴𝐷2 = 9𝐴𝐵2 + 𝐴𝐵2 − 3𝐴𝐵2 ⇒ 9𝐴𝐷2 = 7𝐴𝐵2 Proved 1

27. Cos𝜃 – sin 𝜃 = √2sin𝜃.

Squaring both sides 𝑐𝑜𝑠2𝜃 + 𝑠𝑖𝑛2𝜃 − 2𝑠𝑖𝑛𝜃. 𝑐𝑜𝑠𝜃 = 2𝑠𝑖𝑛2𝜃 1 ⇒ 1 − 2𝑠𝑖𝑛𝜃. 𝑐𝑜𝑠𝜃 = 2 − 2𝑐𝑜𝑠2𝜃 1 ⇒ 1 + 2𝑠𝑖𝑛𝜃. 𝑐𝑜𝑠𝜃 = 2𝑐𝑜𝑠2𝜃 1

⇒ (𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃)2 = (√2𝑐𝑜𝑠𝜃)2

⇒ cos𝜃 + sin 𝜃 = √2𝑐𝑜𝑠𝜃 1

28. LHS. = 𝑡𝑎𝑛𝜃

1−1

𝑡𝑎𝑛𝜃

+1

𝑡𝑎𝑛𝜃

1−𝑡𝑎𝑛𝜃

=𝑡𝑎𝑛2𝜃

𝑡𝑎𝑛𝜃−1−

1

𝑡𝑎𝑛𝜃(𝑡𝑎𝑛𝜃−1)

=𝑡𝑎𝑛3𝜃−1

𝑡𝑎𝑛𝜃(𝑡𝑎𝑛𝜃−1)

=(𝑡𝑎𝑛𝜃−1)(𝑡𝑎𝑛2𝜃+𝑡𝑎𝑛𝜃+1)

𝑡𝑎𝑛𝜃(𝑡𝑎𝑛𝜃−1) 1

= 𝑡𝑎𝑛𝜃 + 1 +1

𝑡𝑎𝑛𝜃

=𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃+ 1 +

𝑐𝑜𝑠𝜃

𝑠𝑖𝑛𝜃

=𝑠𝑖𝑛2𝜃+𝑐𝑜𝑠2𝜃+𝑠𝑖𝑛𝜃.𝑐𝑜𝑠𝜃

𝑐𝑜𝑠𝜃.𝑠𝑖𝑛𝜃 1

=1+𝑠𝑖𝑛𝜃.𝑐𝑜𝑠𝜃

𝑐𝑜𝑠𝜃.𝑠𝑖𝑛𝜃

= 1 + 𝑠𝑒𝑐𝜃. 𝑐𝑜𝑠𝑒𝑐𝜃 1 29.

C.I Fi Xi ui Fiui

0-50 17 25 -2 -34

50-100 35 75 -1 -35

100-150 43 125 0 0

150-200 40 175 1 40

200-250 21 225 2 42

P a g e 63 | 118

250-300 24 275 3 72

∑𝑓𝑖 = 180 ∑𝑓𝑖𝑢𝑖=83 a= 125, h=50 2

x= a + h ∑∫ 𝑖𝑢𝑖

∑∫ 𝑖 1

=125 +50 X 83

180

=148.06 Ans. 1 30. (a) For correct constructing cumulative frequently table. 1

(b) For drawing correct less then type graph. 2 (c) Preventing the child labour. 1

31.

C.I Group A Group B

16-18 50 54

18-20 78 89

20-22 46 40

22-24 28 25

24-26 23 17

For Group A: Modal Class- 18-20 𝑙 = 18, ℎ = 2, 𝑓1 = 78, 𝑓0 = 50, 𝑓2 = 46

Modal age for Group A= 𝑙 + (𝑓1−𝑓0

2𝑓1−𝑓0−𝑓2) 𝑋 ℎ

= 18 + (78−50

156−50−46) 𝑋 2

= 18 +28 𝑋 2

60

= 18.93 For group B:- 𝑙 = 18, ℎ = 2, 𝑓1 = 89, 𝑓0 = 54, 𝑓2 = 40

Modal age of Group B = 18 + (89−50

178−54−40) 𝑋 2

=18.93

P a g e 64 | 118

ACTIVITES (TERM-I)

(Any Eight) Activity 1: To find the HCF of two Numbers Experimentally Based on Euclid Division Lemma

Activity 2: To Draw the Graph of a Quadratic Polynomial and observe:

i. The shape of the curve when the coefficient of x2 is positive

ii. The shape of the curve when the coefficient of x2 is negative

iii. Its number of zero

Activity 3: To obtain the zero of a linear Polynomial Geometrically

Activity 4: To obtain the condition for consistency of system of linear Equations in two variables

Activity 5: To Draw a System of Similar Squares, Using two intersecting Strips with nails

Activity 6: To Draw a System of similar Triangles Using Y shaped Strips with nails

Activity 7: To verify Basic proportionality theorem using parallel line board

Activity 8: To verify the theorem: Ratio of the Areas of Two Similar Triangles is Equal to the Ratio of the Squares of

their corresponding sides through paper cutting.

Activity 9: To verify Pythagoras Theorem by paper cutting, paper folding and adjusting (Arranging)

Activity 10: Verify that two figures (objects) having the same shape (and not necessarily the same size) are similar

figures. Extend the similarity criterion to Triangles.

Activity 11: To find the Average Height (in cm) of students studying in a school.

Activity 12: To Draw a cumulative frequency curve (or an ogive) of less than type.

Activity 13: To Draw a cumulative frequency curve (or an ogive) of more than type.

P a g e 65 | 118

COURSE STRUCTURE (SA-II)

S.NO TOPIC MARKS SA- II

1 ALGEBRA (CONTD.) QUADRATIC EQUATIONS, ARITHMETIC PROGRESSIONS

23

2 GEOMETRY(CONTD.) CIRCLES, CONSTRUCTIONS

17

3 MENSURATION AREAS RELATED TO CIRCLES, SURFACE AREA & VOLUMES

23

4 TRIGONOMETRY(CONTD.) HEIGHT & DISTANCE

8

5 CO-ORDINATE GEOMETRY 11

6 PROBABILITY 8

TOTAL 90

TOPIC WISE ANALYSIS OF EXAMPLES AND QUESTIONS

NCERT TEXT BOOK

CH

AP

TE

RS

TOPICS

Number of Questions for

revision

TOTAL Questions

from

solved

examples

Questions

from exercise

1 QUADRATIC EQUATIONS 18 24 42

2 ARITHMETIC PROGRESSIONS 16 44 60

3 CO-ORDINATE GEOMETRY 15 25 40

4 SOME APPLICATIONS OF

TRIGONOMETRY 7 16 23

5 CIRCLES 3 17 20

6 CONSTRUCTIONS 2 14 16

7 AREA RELATED TO CIRCLES 6 35 41

8 SURFACE AREA & VOLUMES 14 31 45

9 PROBABILITY 13 25 38

Total 94 231 325

P a g e 66 | 118

DETAILS OF THE CONCEPTS TO BE MASTERED BY EVERY CHILD OF CLASS X WITH

EXERCISE AND EXAMPLES OF NCERT TEXT BOOKS.

SA - II

SYMBOLS USED

S. N

o.

TOPIC CONCEPT

DE

GR

EE

OF

IMP

OR

TA

NC

E

DIFFICULTY

LEVEL

NCERT BOOK T.G/L

.G

DE

GR

EE

OF

DIF

FIC

UL

TY

01 Quadratic Equation

Standard form of quadratic equation

* L.G a NCERT Text book

Q.1.2, Ex 4.1

Solution of quadratic equation by factorization

*** L.G a Example 3,4,5, Q.1, 5 Ex. 4.2

Solution of quadratic equation by completing the square

** L.G b Example 8,9 Q.1 Ex. 4.3

Solution of quadratic equation by quadratic formula

*** L.G a Example. 10,11,13,14,15 , Q2,3(ii) Ex.4.3

Nature of roots *** L.G a Example 16 Q.1.2, Ex. 4.4

02 Arithmetic progression

General form of an A.P. * L.G a Exp-1,2, Ex. 5.1 Q.s2(a), 3(a),4(v)

nth term of an A.P. *** L.G a Exp. 3,7,8 Ex. 5.2 Q.4,7,11,16,17,18

Sum of first n terms of an A.P.

** *

** ***

L.G

b

Exp.11,13,15 Ex. 5.3, Q.No.1(i, ii) Q3(i,iii) Q.7,10,12,11,6, Ex5.4, Q-1

03 Coordinate geometry

Distance formula ** L.G b Exercise 7.1, Q.No 1,2,3,4,7,8

Section formula Midpoint formula

**

***

L.G

b

Example No. 6,7,9 Exercise 7.2, Q.No. 1,2,4,5 Example 10. Ex.7.2, 6,8,9. Q.No.7

Area of Triangle **

*** L.G a

Ex.1,2,14 Ex 7.3 QNo-12,4 Ex.7.4, Qno-2

04

Some application of Trigonometry

Heights and distances ** L.G b Example-2,3,4 Ex 9.1 Q 2,5,10,12,13,14,15,16

05 Circles Tangents to a circle *** L.G a Q3(Ex10.1) Q 1,Q6,Q7(Ex 10.2),4

TG/LG is idea identified by term wise error analysis of answers of Q.P. of SA of last three year . * - Important Question a - Low T.G-Teaching Gap ** -Very Important Question b - Average L.G-Learning Gap *** -Very Very Important Question c - Higher

P a g e 67 | 118

Number of tangents from a point to a circle

*** L.G a Theorem 10.1,10.2 Eg 2.1 Q8,9,,10,12,13(Ex 10.2)

06 Constructions

Division of line segment in the given ratio

* L.G b Const 11.1 Ex 11.1 Qno 1

Construction of triangle similar to given triangle as per given scale

*** L.G b Ex 11.1 Qno-2,4,5,7

Construction of tangents to a circle

*** L.G a Ex 11.2 Qno 1,4

07 Area related

to circles

Circumference of a circle * L.G a Example 1 Exercise 12.1 Q.No 1,2,4

Area of a circle * L.G a Example 5,3

Length of an arc of a circle * L.G a Exercise 12.2 Q No 5

Area of sector of a circle ** L.G b Example 2 Exercise 12.2 QNo 1.2

Area of segment of a circle ** L.G a Exercise 12.2 Qno 4,7,9,3

Combination of figures *** L.G b Ex 12.3 Example 4.5 1,4,6,7,9,12,15

08 Surface area and volumes

Surface area of a combination of solids

** T.G c Example 1,2,3 Exercise 13.1 Q1,3,6,7,8

Volume of combination of a solid

** L.G b Example 6 Exercise 13.2 Q 1,2,5,6

Conversion of solids from one shape to another

*** L.G a Example 8 & 10 Exercise 13.3 Q 1,2,6,4,5

Frustum of a cone *** L.G b Example 12& 14 Exercise 13.4 Q 1,3,4,5 Ex-13.5, Q. 5

09 Probability Events * L.G a Ex 15.1 Q4,8,9

Probability lies between 0 and1 ** L.G b Exp- 1,2,4,6,13

Performing experiment *** L.G a Ex 15 1,13,15,18,24

P a g e 68 | 118

QUADRATIC EQUATIONS

KEY POINTS

1. The general form of a quadratic equation is ax2+bx+c=0, a≠o. a, b and c are real numbers.

2. A real number is said to be a root of quadratic equation ax2 + bx + c = 0 where a ≠ 0 if a2 + b + c = 0. The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the corresponding quadratic equation ax2 + bx + c = 0 are the same.

3. Discriminant: - The expression b2-4ac is called discriminan